Question

In following equation calculate the value of H.

1kg steam at$200^{0}C=H+1Kg$ water at ${100}^{0}C$

$\left(S_{steam}=Con\tan t=.5cal/g{m}^{0}C\right)$

Answer 1

In the given equation following processes takes place.

Process 1 : Steam at ${200}^{0}C$ loses out heat to change to steam at ${100}^{0}C$

Heat released in this case = $mc\Delta T$

where, c = specific heat of steam = $0.5cal/g^{0}C$

So, for step 1, heat released = $\left(1000g\right)0.5cal/g^{0}C(200-100{)}^{0}C=50000cal$

Process 2 : Steam at ${100}^{0}C$ gets converted to water at ${100}^{0}C$

This process releases heat. Latent heat of vaporisation of water = $540cal/g^{0}C$

So, 1000 g of steam would release 540000 cal heat.

So, the total heat released in the mentioned equation is (50000+540000) cal = 590000 cal heat = H